1. a. Consider the transformation F(z) = 5z + 4/z acting on the punctured complex plane C- {0}. Find all fixed points of this transformation.
So a fixed point is one where F(z)=z. ie it leaves all points fixed.
So, set f(z) = z ...
F(z) = 5z + 4/z = z
4z = -4/z
4z^2= -4
z^2 = -1
z = i, -i
1. b. Prove that the transformation F preserves the real axis. (i.e., if R, a subset of C, denotes the real axis, prove that F(R) = R).
Define R:
R = { x + 0i is in the complex plane s/t x is a real number, x does not equal zero}
Define F(R)
F(R) = { z is in the complex plane s/t there exists an x that is a real number s/t z = F(x) }
Proof:
Let x be a real number, then F(x) = 5x + (4/x)
By closure of the real numbers, 5x is real, (4/x) is real, and 5x + (4/x) is real, thus F preserves the real axis.
Note: F(R) does not equal 5(R) + (4/R).
1. c.
Prove this does = NOT = preserve y=x. L={x+iy : x,y Is in R, x= y, x not equal to 0}
={x+iy : x is in R, x is not equal to 0}
Provide a counter example. Pick some z=x+xi= re^(i pi/4) r is in R and not equal to 0
Try z=1 + i
F(1+i)= 5(1+i) + 4/(1+i)
=5+5i+ [4/(1+i) * (1-i)/(1-i)]
=5+5i+ (4-4i)/2
=5+5i+2-2
i = 7 +3i Which is not on the line y=x.
OR
Try z=e^(i pi/4)
F(e^(i pi/4)) = 5e^(i pi/4) + 4/e^(i pi/4) = 5e^(i pi/4) + 4e^(-i pi/4)
^^^^not on the line y=x
2. Consider an equilateral triangle T sitting in the complex plan C, centered at the origin. Write down all the transformations of the plane which preserve the triangle. (This means find all transformations of the plane F which satisfy F(T) = T).
(1) A rotation by 120 degrees (2pi/3).
(2) A rotation by 240 degrees (4pi/3).
(3) Identity
(3a) A rotation by 360 degrees (2pi) or 0 degrees (0).
(3b) A translation in which b = 0. T(Triangle ABC) = Triangle ABC + b = Triangle ABC + 0 = Triangle ABC
(4) Reflection
(4a) A reflection over the bisector of angle A.
(4b) A reflection over the bisector of angle B.
(4c) A reflection over the bisector of angle C.
1. a. Consider the transformation F(z) = 5z + 4/z acting on the punctured complex plane C- {0}.
Find all fixed points of this transformation.
So a fixed point is one where F(z)=z. ie it leaves all points fixed.
So, set f(z) = z ...
F(z) = 5z + 4/z = z
4z = -4/z
4z^2= -4
z^2 = -1
z = i, -i
1. b. Prove that the transformation F preserves the real axis. (i.e., if R, a subset of C, denotes the real axis, prove that F(R) = R).
Define R:
R = { x + 0i is in the complex plane s/t x is a real number, x does not equal zero}
Define F(R)
F(R) = { z is in the complex plane s/t there exists an x that is a real number s/t z = F(x) }
Proof:
Let x be a real number, then F(x) = 5x + (4/x)
By closure of the real numbers, 5x is real, (4/x) is real, and 5x + (4/x) is real, thus F preserves the real axis.
Note: F(R) does not equal 5(R) + (4/R).
1. c.
Prove this does = NOT = preserve y=x. L={x+iy : x,y Is in R, x= y, x not equal to 0}
={x+iy : x is in R, x is not equal to 0}
Provide a counter example. Pick some z=x+xi= re^(i pi/4) r is in R and not equal to 0
Try z=1 + i
F(1+i)= 5(1+i) + 4/(1+i)
=5+5i+ [4/(1+i) * (1-i)/(1-i)]
=5+5i+ (4-4i)/2
=5+5i+2-2
i = 7 +3i Which is not on the line y=x.
OR
Try z=e^(i pi/4)F(e^(i pi/4)) = 5e^(i pi/4) + 4/e^(i pi/4) = 5e^(i pi/4) + 4e^(-i pi/4)
^^^^not on the line y=x
2. Consider an equilateral triangle T sitting in the complex plan C, centered at the origin. Write down all the transformations of the plane which preserve the triangle. (This means find all transformations of the plane F which satisfy F(T) = T).
(1) A rotation by 120 degrees (2pi/3).
(2) A rotation by 240 degrees (4pi/3).
(3) Identity
(3a) A rotation by 360 degrees (2pi) or 0 degrees (0).
(3b) A translation in which b = 0. T(Triangle ABC) = Triangle ABC + b = Triangle ABC + 0 = Triangle ABC
(4) Reflection
(4a) A reflection over the bisector of angle A.
(4b) A reflection over the bisector of angle B.
(4c) A reflection over the bisector of angle C.